Under root

Q1. Prove that square root 6 is irrational.

Q2. Prove that cube root 6 is irrational.(without assuming the property that if a prime divides a cube of a no. it divides the no.itself)

#NumberTheory

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Let √6 = p/q (where p and q are integers and are co-prime ) Squaring both sides , 6=p²/q² $\Rightarrow$ p²=6q² (Since , 6 is a factor of p² then it is a factor of p also) Let p=6m (where m is an integer) Squaring both sides , p²=36m² $\Rightarrow$ 6q²=36m² or q²=6m². (Since , p²=6q²) (Since , 6 is a factor of q² then it will be a factor of q also) Hence , 6 is a $**common**$ factor of p and q. This contradicts our fact that p and q are co-prime Hence , √6 cannot be expressed in the form of p/q $\boxed{\therefore \space \sqrt{6} \space is \space irrational}$

Toshit Jain - 4 years, 2 months ago

you may have forgotten that the theorem that you used to prove that 6 is a factor of p applies only for primes !! for example . 12 divides 36 . but 12 does not divide 6

cool math - 4 years, 1 month ago

nice name by the way!!!

Annie Li - 4 years, 1 month ago

cool math - 4 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$